4 Etale algebras, norm and trace

Transcription

1 Number theory I Fall 2017 Lecture #4 09/18/ Etale algebras, norm and trace 4.1 Separability In this section we briefly review some standard facts about separable and inseparable field extensions that we will use repeatedly throughout the course. Those familiar with this material should feel free to skim it. In this section K denotes any field, K is an algebraic closure that we will typically choose to contain any extensions L/K under consideration, and for any polynomial f = a i x i K[x] we use f := iaix i 1 to denote the formal derivative of f (this definition also applies when K is an arbitrary ring). Definition 4.1. A nonzero polynomial f in K[x] is separable if (f, f ) = (1), that is, gcd(f, f ) is a unit in K[x]. Otherwise f is inseparable. If f is separable then it splits into distinct linear factors over over K, where it has deg f distinct roots; this is sometimes used as an alternative definition. Note that the proper of separability is intrinsic to the polynomial f, it does not depend on the field we are working in; in particular, if L/K is any field extension whether or not a polynomial in f K[x] L[x] does not depend on whether we view f as an element of K[x] or L[x]. Warning 4.2. Older texts (such as Bourbaki) define a polynomial in K[x] to be separable if all of its irreducible factors are separable (under our definition); so (x 1) 2 is separable under this older definition, but not under ours. This discrepancy does not change the definition of separable elements or field extensions. Definition 4.3. Let L/K be an algebraic field extension. An element α L is separable over K if it is the root of a separable polynomial in K[x] (in which case its minimal polynomial is necessarily separable). The extension L/K is separable if every α L is separable over K; otherwise it is inseparable. Lemma 4.4. An irreducible polynomial f K[x] is inseparable if and only if f = 0. Proof. Let f K[x] be irreducible; then f is nonzero and not a unit, so deg f > 0. If f = 0 then gcd(f, f ) = f K and f is inseparable. If f is inseparable then g := gcd(f, f ) is a nontrivial divisor of f and f. This implies deg g = deg f, since f is irreducible, but then deg f < deg f = deg g, so g cannot divide f unless f = 0. Corollary 4.5. Let f K[x] be irreducible and let p 0 be the characteristic of K. We n have f(x) = g(x p ) for some irreducible separable g K[x] and integer n 0 that are uniquely determined by f. Proof. If f is separable the theorem holds with g = f and n = 0; for uniqueness, note that if p = 0 then p n n = 0 if and only if n = 0, and if p > 0 and g(x p ) is inseparable unless n = 0 because g(x p n ) = g (x p n )p n n xp 1 = 0 (by the previous lemma). Otherwise f(x) := f r x r is inseparable and f (x) = rfrx r 1 = 0 (by the lemma), and this can occur only if p > 0 and f r = 0 for all r 0 not divisible by p. So f = g(x p ) for some (necessarily irreducible) g K[x]. If g is separable we are done; otherwise we proceed by induction. As above, the n uniqueness of g and n is guaranteed by the fact that g(x p ) = 0 for all n > 0. Corollary 4.6. If char K = 0 then every algebraic extension of K is separable. Andrew V. Sutherland

2 Lemma 4.7. Let L = K(α) be an algebraic field extension contained in an algebraic closure K of K and let f K[x] be the minimal polynomial of α over K. Then # Hom K (L, K) = #{β K : f(β) = 0} [L : K], with equality if and only if α is separable over K. Proof. Each element of Hom K (L, K) is uniquely determined by the image of α, which must be a root β of f(x) in K. The number of these roots is equal to [L : K] = deg f precisely when f, and therefore α, is separable over K. Definition 4.8. Let L/K be a finite extension of fields. The separable degree of L/K is The inseparable degree of f is [L : K] s := # Hom K (L, K). [L : K] i := [L : K]/[L : K] s We will see shortly that [L : K] s always divides [L : K], so [L : K] i is an integer (in fact a power of the characteristic of K), but it follows immediately from our definition that holds regardless. [L : K] = [L : K] s [L : K] i. Theorem 4.9. Let L/K be an algebraic field extension. and let φ K : K Ω be a homomorphism to an algebraically closed field Ω. Then φ K extends to a homomorphism φ L : L Ω. Proof. We use Zorn s lemma. Define a partial ordering on the set F of pairs (F, φ F ) for which F/K is a subextension of L/K and φ F : F Ω extends φ K by defining (F 1, φ F1 ) (F 2, φ F2 ) whenever F 2 con tains F 1 and φ F2 extends φ F1. Given any totally ordered subset C of F, let E be the field {F : (F, φ F ) C} and define φ E : E Ω by φ E (x) = φ F (x) for x F E (this does not depend on the choice of F because C is totally ordered). Then (E, φ E ) is a maximal element of C, and by Zorn s lemma, F contains a maximal element (M, φ M ). We claim that M = L. If not, then pick α L M and consider the field F = M[α] L properly containing M, and extend φ M to ϕ F : F Ω be letting φ F (α) be any root of α M (f) in Ω, where f M[x] is the minimal polynomial of α over M and α M (f) is the image of f in Ω[x] obtained by applying φ M to each coefficient. Then (M, φ M ) is strictly dominated by (F, φ F ), contradicting its maximality. Lemma Let L/F/K be a tower of finite extensions of fields. Then # Hom K (L, K) = # Hom K (F, K)# Hom F (L, K). Proof. We decompose L/F/K into a tower of simple extensions and proceed by induction. The result is trivial if L = K and otherwise it suffices to consider K F F (α) = L, where K = F in the base case. Theorem 4.9 allows us to define a bijection Hom K (F, K) Hom F (F (α), K) Hom K (F (α), K) that sends (φ 1, φ 2 ) to φ: L K defined by φ F = φ 1 and φ(α) = ( ˆφ 1 ˆφ2 ˆφ 1 1 )(α), where ˆφ 1, ˆφ 2 Aut K (K) are arbitrary extensions of φ 1, φ 2 to K; note that φ(α) does not depend on these choices and is a root of φ(f), where f F [x] is the minimal polynomial of α and φ(f) is its image in φ(f )[x]. The inverse bijection is φ 1 = φ F and φ 2 (α) = (φˆ 1 φφ ˆˆ 1 )(α) Fall 2017, Lecture #4, Page 2

3 Corollary Let L/F/K be a tower of finite extensions of fields. Then [L : K] s = [L : F ] s [F : K] s [L : K] i = [L : F ] i [F : K] i Proof. The first equality follows from the lemma and the second follows from the identities [L : K] = [L : F ][F : K] and [L : K] = [L : K] s [L : K] i. Theorem Let L/K be a finite extension of fields. The following are equivalent: (a) L/K is separable; (b) [L : K] s = [L : K]; (c) L = K(α) for some α L separable over K; (d) L K[x]/(f) for some monic irreducible separable polynomial f K[x]. Proof. The equivalence of (c) and (d) is immediate (let f be the minimal polynomial of α and let α be the image of x in K[x]/(f)), and the equivalence of (b) and (c) is given by Lemma 4.7. That (a) implies (c) is the Primitive Element Theorem, see [2, 15.8] or [3, V.7.4] for a proof. It remains only to show that (c) implies (a). So let L = K(α) with α separable over K. For any β L we can write L = K(β)(α), and we note that α is separable over K(β), since its minimal polynomial over K(β) divides it minimal polynomial over K, which is separable. Lemma 4.7 implies [L : K] s = [L : K] and [L : K(β)] s = [L : K(β)] (since L = K(α) = K(β)(α)), and the equalities [L : K] = [L : K(β)][K(β) : K] [L : K] s = [L : K(β)] s [K(β) : K] s then imply [K(β) : K] s = [K(β) : K]. So β is separable over K (by Lemma 4.7). This applies to every β L, so L/K is separable and (a) holds. Corollary Let L/K be a finite extension of fields. Then [L : K] s [L : K] with equality if and only if L/K is separable. Proof. We have already established this for simple extensions, and otherwise we my decompose L/K into a finite tower of simple extensions and proceed by induction on the number of extensions, using the previous two corollaries at each step. Corollary Let L/F/K is a tower of finite extensions of fields. Then L/K is separable if and only if both L/F and F/K separable. Proof. The forward implication is immediate and the reverse implication follows from Corollaries 4.11 and Corollary Let L/F/K be a tower of algebraic field extensions. Then L/K is separable if and only if both L/F and F/K are separable. Proof. As in the previous corollary the forward implication is immediate. To prove the reverse implication, we assume L/F and F/K are separable and show that every β L is separable over K. If β F we are done, and if not we at least know that β is separable over F. Let M/K be the subextension of F/K generated by the coefficients of the minimal polynomial f F [x] of β over F. This is a finite separable extension of K, and M(β) is also a finite separable extension of M, since the minimal polynomial of β over M(β) is f, which is separable. By the previous corollary, M(β), and therefore β, is separable over K Fall 2017, Lecture #4, Page 3

4 Corollary Let L/K be an algebraic field extension, and let Then F is a separable field extension of K. F = {α L : α is separable over K}. Proof. This is clearly a field, since if α and β are both separable over K then K(α) and K(α, β) are separable extensions of K (by the previous corollary), thus every element of K(α, β), including αβ and α + β, is separable over K and lies in F. The field F is then separable by construction. Definition Let L/K be an algebraic field extension. The field F in Corollary 4.16 is the separable closure of K in L. When L is an algebraic closure of K it is simply called a separable closure of K and denoted K sep. When K has characteristic zero the notions of separable closure and algebraic closure necessarily coincide. This holds more generally whenever K is a perfect field. Definition A field K is perfect if every algebraic extension of K is separable. All fields of characteristic zero are perfect. Perfect fields of positive characteristic are characterized by the following property. Theorem A field K of characteristic p > 0 is perfect if and only if K = K p, that is, every element of K is a pth power, equivalently, the map x x p is an automorphism. Proof. If K K p then for any α K K p the polynomial x p α is irreducible and the extension K[x]/(x p α) is inseparable, implying that K is not perfect. Now suppose K = K p n and let f K[x] be irreducible. By Corollary 4.5, we havef(x) = g(x p ) for some separable g K[x] and n 0. If n > 0 then p n p n 1 f(x) = g(x ) = (g (x ) p, where g is the polynomial obtained from g by replacing each coefficient with its pth root (thus g (x) p = g(x p ), since we are in characteristic p). But this contradicts the irreducibility of f. So n = 0 and f = g is separable. The fact that every irreducible polynomial in K[x] is separable implies that every algebraic extension of K is separable, so K is perfect. Corollary Every finite field is a perfect field. n n Proof. If a field K has cardinality p then #K = p 1, thus α = α p = (α p ) p for all α K and every element of K is a pth power. Definition A field K is separably closed if K has no nontrivial finite separable extensions. Equivalently, K is equal to its separable closure in any algebraic closure of K. Definition An algebraic extension L/K is purely inseparable if [L : K] s = 1. Remark The trivial extension K/K is both separable and purely inseparable (but not inseparable!); conversely, an extension that is separable and purely inseparable is trivial. Example If K = F p (t) and L = K[x]/(x p t) = F p (t 1/p ), then L/K is a purely inseparable extension of degree p. n n Fall 2017, Lecture #4, Page 4

5 Proposition Let K be a field of characteristic p > 0. If L/K is purely inseparable of degree p then L = K(a 1/p ) K[x]/(x p a) for some a K K p. Proof. Every α L K is inseparable over K, and by Corollary 4.5 its minimal polynomial over K is of the form f(x) = g(x p ) with f monic. We have 1 < deg f [L : K] = p, so g(x) must be a monic polynomial of degree 1, which we can write as g(x) = x a. Then f(x) = x p a, and we must have a K p since f is irreducible (a difference of pth powers can be factored). We have [L : K(α)] = 1, so L = K(α) K[x]/(x p a) as claimed. Theorem Let L/K be an algebraic extension and let F be the separable closure of K in L. Then L/F is purely inseparable. Proof. If L/K is separable then L = F the theorem holds, so we assume otherwise, in which case the characteristic p of K must be nonzero. Fix an algebraic closure K of K that contains L. Let α L F have minimal polynomial f over F. Use Corollary 4.5 to write n f(x) = g(x p ) with g F [x] irreducible and separable, and n 0. We must have deg g = 1, since otherwise the roots of g would be separable over F, and therefore over K, but not lie in in the separable closure F of K in L. Thus f(x) = x pn a for some a F (since f is monic and deg g = 1). Since we are in characteristic p > 0, we can factor f in F (α)[x] as p n pn n f(x) = x α = (x α) p. There is thus only one F -homomorphism from F (α) to K. The same statement applies to any extension of F obtained by adjoining any set of elements of L (even an infinite set). Therefore # Hom F (L, K) = 1, so [L : F ] s = 1 and L/F is purely inseparable. Corollary Every algebraic extension L/K can be uniquely decomposed into a tower of algebraic extensions L/F/K with F/K separable and L/F purely inseparable. Proof. By Theorem 4.26, we can take F to be the separable closure of K in L, and this is the only possible choice, since we must have [L : F ] s = 1. Corollary The inseparable degree of any finite extension of fields is a power of the characteristic. Proof. This follows from the proof of Theorem Etale algebras We now want to generalize the notion of a separable field extension. By Theorem 4.12, every finite separable extension L/K can be explicitly represented as L = K[x]/(f) for some separable irreducible f K[x]. If f is not irreducible then we no longer have a field, but we do have a ring K[x]/(f) that is also a K vector space, in which the ring multiplication is compatible with scalar multiplication. In other words, L is a (unital) commutative K- algebra whose elements are all separable over K. The notion of separability extends to elements of a K-algebra (even non-commutative ones): an element is separable over K if and only it is the root of some separable polynomial in K[x] (in which case its minimal polynomial must be separable). Recall that the minimal polynomial of an element α of a K-algebra A is the monic generator of the kernel of the K-algebra homomorphism K[x] A defined by x α; note that if A is not a field, minimal polynomials need not be irreducible Fall 2017, Lecture #4, Page 5

6 It follows from the Chinese remainder theorem that if f is separable then the K-algebra K[x]/(f) is isomorphic to a direct product of finite separable extensions of K. Indeed, if f = f 1 f n is the factorization of f into irreducibles in K[x] then K[x] (f) = K[x] (f 1 f n ) K[x] (f 1 ) K[x], (f n ) where the isomorphism is both a ring isomorphism and a K-algebra isomorphism. The separability of f implies that the f i are separable and the ideals (f i ) are pairwise coprime (this justifies our application of the Chinese remainder theorem). We thus obtain a K- algebra that is isomorphic to finite product of separable field extensions K[x]/(f i ) of K. Algebras of this form are called étale algebras (or separable algebras). Definition Let K be a field. An étale K-algebra is a K-algebra L that is isomorphic to a finite product of separable field extensions of K. The dimension of an étale K-algebra is its dimension as a K-vector space. When this dimension is finite we say that L is a finite étale K-algebra. A homomorphism of étale K-algebras is a homomorphism of K-algebras (which means a ring homomorphism that commutes with scalar multiplication). Remark One can define the notion of an étale A-algebra for any noetherian domain A (we will consider this in a later lecture). Example If K is a separably closed field then every étale K-algebra A is isomorphic to K n = K K for some positive integer n (and therefore a finite étale K-algebra). Étale algebras are semisimple algebras. Recall that a (not necessarily commutative) ring R is simple if it is nonzero and has no nonzero proper (two-sided) ideals, and R is semisimple if it is isomorphic to a nonempty finite product of simple rings R i. 1 A commutative ring is simple if and only if it is a field, and semisimple if and only if it is isomorphic to a finite product of fields; this applies in particular to commutative semisimple K-algebras. Every étale K-algebra is thus semisimple (but the converse do es not hold). n The ideals of a semisimple commutative ring R = i=1 R i are easy to describe; each corresponds to a subproduct. To see this, note that the projection maps R R i are surjective homomorphisms onto a simple ring, thus for any R-ideal I, its image in R i is either the zero ideal or the whole ring (note that the image of an ideal under a surjective ring homomorphism is an ideal). In particular, for each index i, either every (r 1,..., r n ) I has r i = 0 or some (r 1,..., r n ) I has r i = 1; it follows that I is isomorphic to the product of the R i for which I projects onto R i. Proposition Let A = K i be a K-algebra written that is a product of field extensions K i /K. Every surjective homomorphism ϕ: A B of K-algebras corresponds to the projection of A on to a subproduct of its factors. Proof. The ideal ker ϕ is a subproduct of K i, thus A ker ϕ im ϕ and B = im ϕ is isomorphic to the complementary subproduct. Proposition 4.32 can be viewed as a generalization of the fact that every surjective homomorphism of fields is an isomorphism. Corollary The decomposition of an étale algebra into field extensions is unique up to permutation and isomorphisms of factors. 1 There are many equivalent (and a few inequivalent) definitions, but this is the simplest Fall 2017, Lecture #4, Page 6

7 Proof. Let A be an étale K-algebra and suppose A is isomorphic (as a K-algebra) to two products of field extensions of K, say m i=1 K i A n Lj. Composing with isomorphisms yields surjective K-algebra homomorphisms π i : L j K i and π j : K i L j. Proposition 4.32 then implies that each K i must be isomorphic to one of the L j and each L j must be isomorphic to one of the K i (and m = n). Our main interest in étale algebras is that they naturally arise from (and are stable under) base change, a notion we now recall. Definition Let ϕ: A B be a homomorphism of rings (so B is an A-module), and let M be any A-module. The tensor product of A-modules M A B is a B-module (with multiplication defined by b(m b ) := m bb ) called the base change (or extension of scalars) of M from A to B. If M is an A-algebra then its base change to B is a B-algebra. We have already seen one example of base change: if M is an A-module and p is a prime ideal of A then M p = M A A p (this is another way to define the localization of a module). Remark Each ϕ: A B determines a functor from the category of A-modules to the category of B-modules via base change. It has an adjoint functor called restriction of scalars that converts a B-module M into an A-module by the rule am = ϕ(a)m (if ϕ is inclusion this amounts to restricting the scalar multiplication by B to the subring A). The ring homomorphism ϕ: A B will often be an inclusion, in which case we have a ring extension B/A (we may also take this view whenever ϕ is injective, which is necessarily the case if A is a field). We are specifically interested in the case where B/A is a field extension and M is a finite étale A-algebra. Proposition Suppose L is a finite étale K-algebra and K /K is any field extension. Then L K K is a finite étale K -algebra of the same dimension as L. Proof. Without loss of generality we assume that L is actually a field; if not L is a product of fields and we can apply the following argument to each of its factors. By Theorem 4.12, L K[x]/(f) for some separable f K[x], and if f = f 1 f 2 f m is the factorization of f in K [x], we have isomorphisms of K -algebras j=1 L K K K [x]/(f) K [x]/(f i ), in which each factor K [x]/(f i ) is a finite separable extension of K (as discussed above, this follows from the CRT because f is separable). Thus L K K is a finite étale K -algebra, and dim K L = deg f = dim K K [x]/(f), so the dimension is preserved. Example Any finite dimensional real vector space V is a finite étale R-algebra (with coordinate-wise multiplication with respect to some basis); the complex vector space V R C is then a finite étale C-algebra of the same dimension. Note that even when an étale K-algebra L is a field, the base change L K K will often not be a field. For example, if K = Q and L Q is a number field, then L K C will never be a field, it will be isomorphic to a C-vector space of dimension [L : K] > 1. i Fall 2017, Lecture #4, Page 7

8 Remark In the proof of Proposition 4.36 we made essential use of the fact that the elements of an étale K-algebra are separable. Indeed, the proposition does not hold if L is a finite semisimple commutative K-algebra that contains an inseparable element. Corollary Let L K[x]/(f) be a finite separable extension of a field K defined by an irreducible separable polynomial f K[x]. Let K /K be any field extension, and let f = f 1 f m be the factorization of f into distinct irreducible polynomials f i K [x]. We have an isomorphism of finite étale K -algebras L K K K [x]/(f i ) where each K [x]/(f i ) is a finite separable field extension of K. Proof. This follows directly from the proof of Proposition The following proposition gives several equivalent characterizations of finite étale algebras, including a converse to Corollary 4.39 (provided the field K is not too small). Recall that an element α of a ring is nilpotent if α n = 0 for some n, and a ring is reduced if it contains no nonzero nilpotents. Theorem Let L be a commutative K-algebra of finite dimension and assume that the dimension of L is less than the cardinality of K. The following are equivalent: (a) L is a finite étale K-algebra. (b) Every element of L is separable over K. (c) L K K is reduced for every extension K /K. (d) L K K is semisimple for every extension K /K. (e) L = K[x]/(f) for some separable f K[x]. The implications (a) (b) (c) (d) (e) hold regardless of the dimension of L. n Proof. To show (a) (b), let L = i=1 K i with each K i /K separable, and consider n α = (α 1,..., α n ) L = i=1 K i. Each α i K i is separable over K with separable minimal polynomial f i K[x], and α is a root of f := lcm{f 1,..., f n }, which is separable (the LCM of a finite set of separable polynomials is separable), thus α is separable. To show (b) (c), note that if α L is nonzero and separable over K it cannot be nilpotent (the minimal polynomial of a nonzero nilpotent is x n for some n > 1 and is therefore not separable), and separability is preserved under base change. The equivalence (c) (d) follows from Lemma 4.42 below. To show (d) (a), we first note we can assume L is semisimple (take K = K), and it suffices to treat the case where L is a field. By base-changing to the separable closure of K in L, we can further reduce to the case that L/K is a purely inseparable field extension. If L = K we are done. Otherwise we may pick an inseparable α L, and, as in the proof of Theorem 4.26, the minimal polynomial of α has the form f(x) = x pn a for some a K and n 1. Now consider γ := α 1 1 α L K L n We have γ 0, since γ / K, but γ p = α pn 1 1 α pn = a 1 1 a = 0, so γ is a nonzero nilpotent and L K L is not reduced, contradicting (c) (d). i Fall 2017, Lecture #4, Page 8

9 n We have (e) (a) form Corollary For the converse, suppose L = i=1 L i with each L i /K a finite separable extension of K. Pick a monic irreducible separable polynomial f 1 (x) so that L 1 K[x]/(f 1 (x)), and then do the same for i = 2,..., n ensuring that each polynomial f j we pick is not equal to f i for any i < j. This can be achieved by replacing f j (x) with f j (x + a) for some a K if necessary. Here we use the fact that there are at least n distinct choices for a, under our assumption that the dimension of L is less than the cardinality of K (note that if f(x) is irreducible then the polynomials f(x + a) are irreducible and pairwise coprime as a ranges over K). The polynomials f 1,... f n are then coprime and separable, so their product f is separable and L = K[x]/(f), as desired. Remark K-algebras of the form L = K[x]/(f(x)) are monogenic (generated by one element). Theorem?? implies that finite étale K-algebras are monogenic whenever the base field K is big enough. This always holds if K is infinite, but if K is a finite field then not every finite étale K-algebra is monogonic. The recent preprint [5] gives exact bounds on the maximal number of generators needed for a finite étale K-algebra over a finite field. The following lemma is a standard exercise in commutative algebra that we include for the sake of completeness. Lemma Let K be a field. A commutative K-algebra of finite dimension is semisimple if and only if it is reduced. Proof. If A is semisimple it is clearly reduced (otherwise we could project a nonzero nilpotent of A to a nonzero nilpotent in a field); we only need to prove the converse. Every ideal of a commutative K-algebra A is also a K-vector space; this implies that when dim K A is finite A satisfies both the ascending and descending chain conditions and is therefore neotherian and artinian. This implies that A has finitely many maximal ideals M 1,... M n and that the intersection of these ideals (the radical of A) is equal to the set of nilpotent elements of A (the nilradical of A); see Exercises and in [1], for example. Taking the product of the projection maps A A/M i yields a surjective ring hon momorphism ϕ: A i=1 A/M i from A to a product of fields. If A is reduced then ker ϕ = M i = {0} and ϕ is an isomorphism, implying that A is semismiple. Proposition Suppose L is a finite étale K-algebra and Ω is a separably closed field extension of K. There is an isomorphism of finite étale Ω-algebras L K Ω Ω σ HomK(L,Ω) that sends β 1 to the vector (σ(β)) σ for each β L. Proof. We may reduce to the case that L = K[x]/(f) is a separable field extension, and we may then factor f(x) = (x α 1 ) (x α n ) over Ω, with the α i are distinct. We have a bijection between Hom K (K[x]/(f), Ω) and the set {α i }: each σ Hom K (K[x]/(f), Ω) is determined by σ(x) {α i }, and for each α i, the map x α i determines a K-algebra homomorphism σ i Hom K (K[x]/(f), Ω). As in the proof of Proposition 4.36 we have Ω-algebra isomorphisms K[x] Ω[ K (f) Ω x] (f) n Ω[x] (x α i ) i=1 n Ω. i= Fall 2017, Lecture #4, Page 9

10 which map x 1 x (α 1,..., α n ) (σ 1 (x),..., σ n (x)). The element x 1 generates L K Ω as an Ω-algebra, and it follows that β 1 (σ(β)) σ for every β L. Remark The proof of Proposition 4.43 does not require Ω to be separably closed, we could replace Ω with the comp ositum of the normal closure of the field extensions L i/k in the decomposition of L = L i into separable extensions of K (in the proof above we just needed f to split into linear factors). Example Let L/K = Q(i)/Q and Ω = C. We have Q(i) Q[x]/(x 2 + 1) and Q[x] Q(i) Q C x Q C C[x] (x 2 + 1) C[x] (x i) C[x] C (x + i) C. As C-algebra isomorphisms, the corresponding maps are determined by i 1 x 1 x (x, x) (i, i) (i, i). Taking the base change of Q(i) to C lets us see the two distinct embeddings of Q(i) in C, which are determined by the image of i. Note that Q(i) is canonically embedded in its base change Q(i) Q C to C via α α 1. We have 1 = i 2 = (i 1) 2 = i = 1 1 = (1 1) Thus as an ( isomorphism ) of C-algebras, the basis (1 1, i 1) for Q(i) Q C is mapped to the basis (1, 1), (i, i) for C C. For any (α, β) C C, the inverse image of in Q(i) C under this isomorphism is α + β 2 α + β (α, β) = (1, 1) + α β (i, i) 2 2i (1 1) + α β (i 1) = 1 α + β + i α β. 2i 2 2i Now R/Q is an extension of rings, so we can also consider the base change of the Q-algebra Q(i) to R. But note that R is not separably closed and in particular, it does not contain a subfield isomorphic to Q(i), thus Proposition 4.43 does not apply. Indeed, as an R-module, we have Q(i) Q R R 2, but as an R-algebra, Q(i) 2 Q R C R. 4.3 Norms and traces We now introduce the norm and trace map associated to a finite free ring extension B/A. These are often defined only for field extensions, but in fact the same definition works without modification whenever B is a free A-module of finite rank. One can generalize further to projective modules (with some restrictions), but we will not need this. Definition Let B/A be a (commutative) ring extension in which B is a free A-module of finite rank. The (relative) norm N B/A (b) and trace T B/A (b) of b (down to A) are the determinant and trace of the A-linear multiplication-by-b map B B defined by x bx Fall 2017, Lecture #4, Page 10

11 As a special case, note that if A is a field and B is a finite A-algebra (a field extension, for example) then B is an A-vector space of finite dimension, hence a free A-module of finite rank. In practice one computes the norm and trace by picking a basis for B as an A-module and computing the matrix of the multiplication-by-b map with respect to this basis; this is an n n matrix with entries in A whose determinant and trace are basis independent. It follows immediately from the definition that N B/A is multiplicative, T B/A is additive, we have group homomorphisms N B/A : B A and T B/A : B A, and if B 1 /A and B 2 /A are two ring extensions that are free A-modules of finite rank then N B1 B 2 /A(x) = N B1 /A(x 1 )N B2 /A(x 2 ) and T B1 B 2 /A = T B1 /A(x 1 ) + TB 2 /A(x 2 ) for all x = (x 1, x 2 ) B 1 B 2. Example Consider A = R and B = C, which has the A-module basis (1, i). For b = 2 + 3i the matrix of B b ( B with respect to this basis can be written as 2 3 ) 3 2, thus ( ) 2 3 N C/R(2 + 3i) = det = 13, 3 2 ( ) 2 3 T C / R(2 + 3i) = tr = Warning In order to write down the matrix of an A-linear transformation B B with respect to basis for B as a free A-module of rank n, we not only need to pick a basis, we need to decide whether to represent elements of B A n as row vectors with linear transformations acting via matrix multiplication on the right, or as column vectors with linear transformations acting via matrix multiplication on the left. The latter convention is often implicitly assumed in the literature (as in the example above), but the former is often used in computer algebra systems (such as Magma). We now verify that the norm and trace are well behaved under base change. Lemma Let B/A be ring extension with B free of rank n over A, and let ϕ: A A be a ring homomorphism. The base change B = B A A of B to A is a free A -module of rank n and we for every b B ϕ(n B/A (b)) = N B /A (b 1) and ϕ(t B/A (b) = TB /A (b 1). Proof. Let b B, let (b 1,..., b n ) be a basis for B as an A-module, and let M = (m ij ) A n n b be the matrix of B B with respect to this basis. Then (b 1 1,..., b n 1) is a basis for B as an A -module (thus B is free of rank n over A ) and M = (ϕ(m ij)) A n n is the matrix of B b 1 B, and we have ϕ(n B/A (b)) = ϕ(det M) = det M = N B /A (b 1) ϕ(t B/A (b)) = ϕ(tr M) = tr M = N B /A (b 1) Theorem Let K be a field with separable closure Ω and let L be a finite étale K- algebra. For all α L we have N L/K (α) = σ(α) and TL/K (α) = σ(α). σ Hom K (L,Ω) σ Hom K (L,Ω) Fall 2017, Lecture #4, Page 11

12 Proof. Let n be the rank of L as a K-module. By the previous lemma and Proposition 4.43, N L/K (α) = N (L K Ω)/Ω(α 1) = N Ω n /Ω((σ 1 (α),..., σ n (α)) = n σi (α). The isomorphism L K Ω σ Ω = Ωn of Prop sends α 1 to (σ 1 (α),..., σ n (α)). Using the standard basis for Ω n, the matrix of multiplication-by-(σ 1 (α),..., σ n (α)) is just the diagonal matrix with σ i (α) in the ith diagonal entry. Similarly, T L/K (α) = T (L K Ω)/Ω(α 1) = T Ω n /Ω((σ 1 (α),..., σ n (α)) = i=1 n σi (α). The proof above demonstrates a useful trick: when working over a field that is not algebraically/separably closed, base change to an algebraic/separable closure. This often turns separable field extensions into étale algebras that are no longer fields. Proposition Let L/K be a (not necessarily separable) finite extension, let K be an algebraic closure of K containing L. Let α L have minimal polynomial f K[x] with d factorization f(x) = i=1 (x α i ) in K[x], and let e = [L : K(α)]. We have N L/K (α) = d i=1 α e i and T L/K (α) = e d αi. d In particular, if f(x) = i=0 a ix i, then N L/K (α) = ( 1) de a e 0 and T L/K(α) = ea d 1. Proof. See Problem Set 2. Corollary Let M/L/K be a tower of finite extensions. Then N M/K = N L/K N M/L and T M/K = T L/K T M/L. Proof. Fix a separable closure Ω of K that contains M. As in the proof of Lemma 4.10, each σ Hom K (M, Ω) can be identified with a pair (σ 1, σ 2 ) with σ 1 Hom L (M, Ω) and σ 2 Hom K (L, Ω). We then note that for any α M, N M/K (α) = σ(α) = σ 2 σ 1 (α) = N L/K (N M/L (α)), σ Hom K (M,Ω) σ 2 Hom K (L,Ω) σ 1 HomL(M,Ω) and T M/K (α) = T L/K (T M/L( α)) follows similarly by replacing products with sums. Corollary Let A be an integrally closed domain with fraction field K and let L/K be a finite extension. if α L is integral over A then N L/K (α) A and T L/K (α) A. Proof. This follows immediately from Propositions 1.27 and Corollary 4.52 actually holds in much greater generality. Theorem 4.54 (Transitivity of Norm and Trace). Let A B C be rings with C free of finite rank over B and B free of finite rank over A. Then C is free of finite rank over A and N C/A = N B/A N C/B and T C/A = T B/A T C/B. Proof. See [3, III.9.4]. i=1 i= Fall 2017, Lecture #4, Page 12

13 References [1] Allen Altman and Steven Kleiman, A term of commutative algebra, Worldwide Center of Mathematics, [2] Michael Artin, Algebra, 2nd edition, Pearson, [3] Nicolas Bourbaki, Algebra I: Chapters 1 3, Springer, [4] Nicolas Bourbaki, Algebra II: Chapters 4 7, Springer, [5] Uriya First, Zinovy Reichstein, Santiago Salazar, On the number of generators of a separable algebra over a finite field, arxiv: , [6] Anthony W. Knapp, Advanced Algebra, Digital Second Edition, [7] Joseph J. Rotman, Advanced Modern Algebra, 2nd edition, Graduate Studies in Mathematics 114, AMS, Fall 2017, Lecture #4, Page 13

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